#Instantaneous Rate of Change

#Table of Contents

1. What is the Instantaneous Rate of Change?
2. Finding the Instantaneous Rate of Change
3. Worked Example
4. Practice Questions
5. Glossary
6. Summary and Key Takeaways

#What is the Instantaneous Rate of Change?

The instantaneous rate of change is the slope of a graph at a specific point, rather than between two points.

#Finding the Instantaneous Rate of Change

Consider the graph of a function $f$ with two points on the graph, $A$ and $B$:

#Average Rate of Change

Using the labeling on the left:

• The average rate of change from $A$ to $B$ can be written as: $$\frac{f(x) - f(a)}{x - a}$$

Using the labeling on the right:

• The average rate of change from $A$ to $B$ can be written as: $$\frac{f(a + h) - f(a)}{h}$$

#Instantaneous Rate of Change

To find the instantaneous rate of change, consider finding the slope between $A$ and $A$:

• As the distance between the two points becomes smaller, the slope will become a more accurate estimate for the instantaneous rate of change at $x = a$.
• Therefore, the instantaneous rate of change will be the limit as this distance tends to zero.

The instantaneous rate of change at $x = a$ can be written as: $$\underset{x \to a}{\mathrm{lim}} \frac{f(x) - f(a)}{x - a}$$ or $$\underset{h \to 0}{\mathrm{lim}} \frac{f(a + h) - f(a)}{h}$$

These only give a valid answer if the relevant limit exists. They are both equivalent forms of the definition of the derivative of the function at $x = a$, denoted by $f'(a)$.

#Worked Example

A function $f(x)$ is defined by $f(x) = x^2$. Using the equation below, find the instantaneous rate of change of $f(x)$ at the point where $x = 2$.

$$f'(x) = \underset{h \to 0}{\mathrm{lim}} \frac{f(x + h) - f(x)}{h}$$

Solution:

1. Substitute $x = 2$ into the given formula: $$f'(2) = \underset{h \to 0}{\mathrm{lim}} \frac{f(2 + h) - f(2)}{h}$$

2. Evaluate the function, $f(x) = x^2$, at $x = 2$: $$f'(2) = \underset{h \to 0}{\mathrm{lim}} \frac{(2 + h)^2 - 2^2}{h}$$

3. Expand and simplify the numerator: $$f'(2) = \underset{h \to 0}{\mathrm{lim}} \frac{4 + 4h + h^2 - 4}{h}$$ $$f'(2) = \underset{h \to 0}{\mathrm{lim}} \frac{4h + h^2}{h}$$

4. Simplify the fraction by canceling terms: $$f'(2) = \underset{h \to 0}{\mathrm{lim}} (4 + h)$$

5. Evaluate the limit: $$f'(2) = 4$$

Therefore, the instantaneous rate of change of $f(x)$ at $x = 2$ is 4.

#Practice Questions

Practice Question

1. Find the instantaneous rate of change of $f(x) = x^3$ at $x = 1$.

Practice Question

2. Determine the instantaneous rate of change of $f(x) = \sin(x)$ at $x = \frac{\pi}{4}$.

Practice Question

3. Calculate the instantaneous rate of change of $f(x) = e^x$ at $x = 0$.

#Glossary

• Instantaneous Rate of Change: The rate at which a function is changing at any given point.
• Derivative: A measure of how a function changes as its input changes.
• Limit: A value that a function approaches as the input approaches some value.

#Summary and Key Takeaways

• The instantaneous rate of change is the slope of a graph at a specific point.
• It is found by taking the limit of the average rate of change as the distance between two points approaches zero.
• The derivative of a function at a point provides the instantaneous rate of change at that point.
• Practice simplifying expressions before evaluating limits to avoid errors.

The instantaneous rate of change has numerous applications in real-world scenarios, such as calculating speed in physics and rates of reaction in chemistry.